In Chemistry we measure quantities in 2 main ways. We measure a chemical in grams by determining its molecular mass. We measure a chemical in moles also.
The big numbers in a chemical equation (the ones in front of a compound) represents the number of moles required for the reaction to go to completion.
We can convert the quantities of moles to grams and vice versa. Use this relationship number of moles = the mass I have / the molar mass
n = m/M
When we use Avagadro's number we are measuring the number of particles in a mole. number moles x 6.02 x10 23 = number of particles (I have or in the question) n = number of particles I have / number of particles in a mole ( ie Avagadro's number)
(RMM, symbol, Mr) is the total of the Relative Atomic Mass of each atom in a compound. There is no unit for Mr. If a dot (.) is in the formula, the atomic masses are added not multiplied. eg. Mr (CuSO4.5H2O) = 63.5 (Cu) + 32 (S) + 4 x 16 (O) + 5 x [2 x 1 (H) + 16 (O)] = 249.5.
For ionic compounds that are made up of many units of the cation and anion, the Molecular Mass is called the Formula Mass and is equal to the Mr of the ionic compound’s formula. eg. Mr(NaCl) = 23 + 35.5 = 58.8
The Molar Mass (symbol, M) is the mass of 1 mole of any substance and is numerically equal to the Relative Molecular Mass but has a unit _. Therefore the molar mass of CuSO4.5H2O = 249.5 g mol-1.
Measuring the Amount of Different Substances: The Mole (symbol, n and unit, mol) is the Amount Of Substance containing a fixed number of particles and is the standard way Chemists describe the amount of substance in different states (solid, solution, liquid or gas). This is not a concentration. Concentration is the number of moles per litre - and we'll deal with that when we study acids and bases.
The number of particles in 1 mole of any substance is equal to the value of Avogadro's Constant (symbol Na) which is 6.02 x 1023mol-1. This value is also equal to the number of particles in _ of Carbon. To use the different mol formulae correctly, the correct units for each substance’s measure must be used:- Mass, m is in grams, g. (1000 mg = 1.0 g, 1.0 kg = 1000 g and 1 tonne = 1000 kg). Volume, V is in Litres, L. (1000 mL (cm3) = 1.0 L (dm3) and 1 m3= 1000 dm3). Concentration, c is in Molarity, M or mol/L. Pressure, P is in kiloPascals, kPa. (101.325 kPa = 101325 Pa = 760 mmHg = 1.0 atmosphere pressure). Temperature, T is on Kelvin, K (Temperature in Kelvin = Temperature in degrees Celsius, oC + 273 K). For particles (atoms, molecules, ions), use, the amount, n in mol = For masses of solids, liquids or gases, use, the amount, n in mol = For _ (not pure liquids), use, the amount, n in mol = Concentration, c in M x Volume, V in L Covered later For gases, use, the amount, n in mol = covered later
In calculations, the final answer is written with the same number of significant figures as the value with the least number of significant figures used in the calculation. Any 0 before any digit (including a decimal point) is not a significant figure but all digits including & after the first non-zero digit are counted as significant figures. eg. 0.12 has 2 sig. figures, 1.02 has 3 sig. figures, 20.00 has 4 sig. figures and 0.002 has 1 sig. figure (2 x 10-3). If values are very low or high use Standard Form. eg. 30257 g = 3.0 x 104 g and for 0.003 g = 3.0 x 10-3 g. When doing mol calculations, ensure the Chemical is shown in a ( ) after writing n. eg. n(Chemical) = Basic question to answer when doing calcuations - “How many moles is that”
Determining the Amount in Mol
Determining another unit using the Amount in mol
For Particles
For Particles
1. Determine the amount (in mol) present in 1.20 x 1024 molecules of H2O. Number of particles = 1.20 x 1024, Find n
n(H2O)= 1.20 x 10 24 / 6.02 x 10 23
n(H2O)= 2.00 mol
1. Determine the number of atoms in 2.0 mol of H2O. Number of particles = n x Avogadros Constant. n = 2.0, Find the number of particles
Number of H2O= n(H2O) x 6.02 x 1023. = 2.00 x 6.02 x 1023 = 1.20 x 1024 H2O molecules. 1 H2O molecule has 2 x H + 1 x O = 3 atoms Number of atoms = 3 x 1.2 x 1024 = 3.6 x 1024 atoms
For Masses of solid/liquid/gas solutes
For Masses of solid/liquid/gas solutes
2. Determine the amount (in mol) present in 12 g of Oxygen. m = 12 g, Find n Oxygen like most non metals (not Noble gases) exists as diatomic molecules (2 atoms) = O2. n(O2) = = n(O2) = 0.38 mol (not 0.375) due to sig figs
2. Determine the mass of Magnesium present in 3.3 mol of Magnesium. n = 3.3, Find the mass Magnesium like all metals, exists as single atoms = Mg.
m(Mg) = n(Mg) x Mr(Mg), The Mr are on Pg 313. = m(Mg) =
For solutes in Solutions
For solutes in Solutions
3. Determine the amount (in mol) of NaCl present in 500 mL of 0.25 M NaCl solution.
V = 500 mL = 0.500 L (mL ÷ 1000 = L), c = 0.25 M, Find n
n(NaCl) = Conc. c in M x Volume, V in L = n(NaCl) =
3. Determine the concentration of NaCl present in a 2.0 L solution containing 0.12 mol of NaCl. n = 0.12, V = 2.0 L, Find c [NaCl] = The [ ] represents the conc. = [NaCl] = 0.060 M.
Some calculations involve the combination of 2 or more Mol formula to determine a measurement. Use the measurements stated to determine the amount in mol (n) and then use this to find the required measurement.
Determine the mass of Glucose, C6H12O6 present in 2.0 L of 0.25 M Glucose solution
V = 2.0 L, c = 0.25 M, Find n and then find m
n(C6H12O6) = c x V = 0.25 x 2.0 = 0.50 mol m(C6H12O6) = n(C6H12O6) x Mr = 0.50 x 180 m(C6H12O6) = 90 g.
Determine the number of Glucose, C6H12O6 molecules present in 2.0 L of 0.50 M Glucose solution V = 2.0 L, c = 0.25 M, Find n and then find number of molecules n(C6H12O6) = c x V = 0.50 x 2.0 = 1.0 mol No. of C6H12O6= n(C6H12O6) x Na (6.02 x 1023) = 1 x 6.02 x 1023 = 6.0 x 1023 C6H12O6 molecules
Empirical formula calculations
Facts: an empirical formula is like the lowest common denominator. Eg CH3 is the empirical formula C2H6
is not - it is a molecular formula and as such has a specific mass. (in this case 30 g 2 x 12 + 1 x 6) C3H9 is also a molecular formula and has a mass of 45g (12 x 3 +1 x 9)
The other thing the empirical formula tells us is the ratio in which the atoms of the elements combine. For our example above CH3 it is 1 : 3, (C : H). It also tells us the atoms of the elements combine in the molar ratio of 1:3.
The molecular formula is dependent on the compounds molecular mass. For example if a subtance has a molecular mass of 45g and a formula mass ( based on its empirical formula) of 15g then we can predict that there are 3 empirical units in this molecule. and hence the molecular formula will be
3 x CH3 = C3H9
Example
A compound contains 27.3% C and 72.7% oxygen. Calculate the empirical formula.
Carbon
Oxygen
Step 1.
Record the mass in grams. (m)
If % is given assume you have 100g
27.3
72.7
Step 2.
Calculate the amount in moles (n)
27.3/12.0
=2.27
72.7 / 16.0
=4.54
Step 3.
Divide each of the moles by the smallest mole value in step 2
2.27 /2.27
=1
4.54 /2.27
=2
Step 4
Obtain the simplest whole number mole ratio
1
2
therefore the empirical formula CO2
Example 2
A sample of a compound of aluminium and oxygen has a mass of 2.36g. It contains 1.25g Al. Find the empirical formula
Al
O
Step 1
m(g)
1.25
2.36-1.25=1.11
Step 2
n (mol)
1.25 / 27.0=
0.0463
1.11 / 16.0 =
0.0694
Step 3
Divide all amounts by the smallest number of moles
0.0463 /0.0463 =
1.0
0.0694 / 0.0463 =
1.5
Step 4
Multiply both numbers by a factor that makes all the numbers in step 3 whole numbers ( in this case 2)
2
3
Therefore the empirical formula is Al2O3
Calculating Molecular formula
Overview - calculate the empirical formula, then use the mass given to identify the number of moles
Example
Hydrocarbon contains 7.2g of C and 1.5g of H. The molar mass of the compound was 58 g mol -1.
C
H
Step 1
m (g)
7.2
1.5
Step 2
n (mol)
7.2 / 12
= 0.6
1.5 / 1.0
= 1.5
Step 3
Divide all amounts by the smallest number of moles from Step 2
0.6 / 0.6
= 1
1.5 / 0.6
= 2.5
Step 4
Give simplest whole number mole ratio (x 2)
2
5
So the empirical formula is C2H5 the molecular mass of C2H5 is 12 x 2 + 1 x 5 = 29g But the molar mass of the compound is 58g ( data given from the question) Number of C2H5 s in one of these 58g moles is 58 / 29 = 2 therefore the molecuar formula is 2 x the empirical formula ie C4H10
Chap 4. Summary Moles and Masses
In Chemistry we measure quantities in 2 main ways. We measure a chemical in grams by determining its molecular mass. We measure a chemical in moles also.The big numbers in a chemical equation (the ones in front of a compound) represents the number of moles required for the reaction to go to completion.
We can convert the quantities of moles to grams and vice versa. Use this relationship
number of moles = the mass I have / the molar mass
n = m/M
When we use Avagadro's number we are measuring the number of particles in a mole.
number moles x 6.02 x10 23 = number of particles (I have or in the question)
n = number of particles I have / number of particles in a mole ( ie Avagadro's number)
n = N / Na
The Relative Molecular Mass
Simulation -
http://phet.colorado.edu/en/simulation/isotopes-and-atomic-mass
(RMM, symbol, Mr) is the total of the Relative Atomic Mass of each atom in a compound. There is no unit for Mr. If a dot (.) is in the formula, the atomic masses are added not multiplied.
eg. Mr (CuSO4.5H2O) = 63.5 (Cu) + 32 (S) + 4 x 16 (O) + 5 x [2 x 1 (H) + 16 (O)] = 249.5.
For ionic compounds that are made up of many units of the cation and anion, the Molecular Mass is called the Formula Mass and is equal to the Mr of the ionic compound’s formula. eg. Mr(NaCl) = 23 + 35.5 = 58.8
The Molar Mass (symbol, M) is the mass of 1 mole of any substance and is numerically equal to the Relative Molecular Mass but has a unit _. Therefore the molar mass of CuSO4.5H2O = 249.5 g mol-1.
Measuring the Amount of Different Substances:
The Mole (symbol, n and unit, mol) is the Amount Of Substance containing a fixed number of particles and is the standard way Chemists describe the amount of substance in different states (solid, solution, liquid or gas).
This is not a concentration. Concentration is the number of moles per litre - and we'll deal with that when we study acids and bases.
The number of particles in 1 mole of any substance is equal to the value of Avogadro's Constant (symbol Na) which is 6.02 x 1023 mol-1. This value is also equal to the number of particles in _ of Carbon.
To use the different mol formulae correctly, the correct units for each substance’s measure must be used:-
Mass, m is in grams, g. (1000 mg = 1.0 g, 1.0 kg = 1000 g and 1 tonne = 1000 kg).
Volume, V is in Litres, L. (1000 mL (cm3) = 1.0 L (dm3) and 1 m3= 1000 dm3).
Concentration, c is in Molarity, M or mol/L.
Pressure, P is in kiloPascals, kPa. (101.325 kPa = 101325 Pa = 760 mmHg = 1.0 atmosphere pressure).
Temperature, T is on Kelvin, K (Temperature in Kelvin = Temperature in degrees Celsius, oC + 273 K).
For particles (atoms, molecules, ions), use, the amount, n in mol =
For masses of solids, liquids or gases, use, the amount, n in mol =
For _ (not pure liquids), use, the amount, n in mol = Concentration, c in M x Volume, V in L Covered later
For gases, use, the amount, n in mol = covered later
In calculations, the final answer is written with the same number of significant figures as the value with the least number of significant figures used in the calculation. Any 0 before any digit (including a decimal point) is not a significant figure but all digits including & after the first non-zero digit are counted as significant figures.
eg. 0.12 has 2 sig. figures, 1.02 has 3 sig. figures, 20.00 has 4 sig. figures and 0.002 has 1 sig. figure (2 x 10-3).
If values are very low or high use Standard Form. eg. 30257 g = 3.0 x 104 g and for 0.003 g = 3.0 x 10-3 g.
When doing mol calculations, ensure the Chemical is shown in a ( ) after writing n. eg. n(Chemical) =
Basic question to answer when doing calcuations - “How many moles is that”
Determining the Amount in Mol
Determining another unit using the Amount in mol
For Particles
For Particles
1.20 x 1024 molecules of H2O.
Number of particles = 1.20 x 1024, Find n
n(H2O)= 1.20 x 10 24 / 6.02 x 10 23
n(H2O)= 2.00 mol
Number of particles = n x Avogadros Constant.
n = 2.0, Find the number of particles
Number of H2O= n(H2O) x 6.02 x 1023.
= 2.00 x 6.02 x 1023
= 1.20 x 1024 H2O molecules.
1 H2O molecule has 2 x H + 1 x O = 3 atoms
Number of atoms = 3 x 1.2 x 1024 = 3.6 x 1024 atoms
For Masses of solid/liquid/gas solutes
For Masses of solid/liquid/gas solutes
12 g of Oxygen.
m = 12 g, Find n
Oxygen like most non metals (not Noble gases) exists as diatomic molecules (2 atoms) = O2.
n(O2) =
=
n(O2) = 0.38 mol (not 0.375) due to sig figs
Magnesium.
n = 3.3, Find the mass
Magnesium like all metals, exists as single atoms = Mg.
m(Mg) = n(Mg) x Mr(Mg), The Mr are on Pg 313.
=
m(Mg) =
For solutes in Solutions
For solutes in Solutions
present in 500 mL of 0.25 M NaCl solution.
V = 500 mL = 0.500 L (mL ÷ 1000 = L), c = 0.25 M, Find n
n(NaCl) = Conc. c in M x Volume, V in L=
n(NaCl) =
solution containing 0.12 mol of NaCl.
n = 0.12, V = 2.0 L, Find c
[NaCl] = The [ ] represents the conc.
=
[NaCl] = 0.060 M.
in 2.0 L of 0.25 M Glucose solution
V = 2.0 L, c = 0.25 M, Find n and then find m
n(C6H12O6) = c x V= 0.25 x 2.0
= 0.50 mol
m(C6H12O6) = n(C6H12O6) x Mr
= 0.50 x 180
m(C6H12O6) = 90 g.
in 2.0 L of 0.50 M Glucose solution
V = 2.0 L, c = 0.25 M, Find n and then find number of molecules
n(C6H12O6) = c x V
= 0.50 x 2.0
= 1.0 mol
No. of C6H12O6= n(C6H12O6) x Na (6.02 x 1023)
= 1 x 6.02 x 1023
= 6.0 x 1023 C6H12O6 molecules
Empirical formula calculations
Facts: an empirical formula is like the lowest common denominator. Eg CH3 is the empirical formula C2H6is not - it is a molecular formula and as such has a specific mass. (in this case 30 g 2 x 12 + 1 x 6) C3H9 is also a molecular formula and has a mass of 45g (12 x 3 +1 x 9)
The other thing the empirical formula tells us is the ratio in which the atoms of the elements combine. For our example above CH3 it is 1 : 3, (C : H). It also tells us the atoms of the elements combine in the molar ratio of 1:3.
The molecular formula is dependent on the compounds molecular mass. For example if a subtance has a molecular mass of 45g and a formula mass ( based on its empirical formula) of 15g then we can predict that there are 3 empirical units in this molecule. and hence the molecular formula will be
3 x CH3 = C3H9
Example
A compound contains 27.3% C and 72.7% oxygen. Calculate the empirical formula.Record the mass in grams. (m)
If % is given assume you have 100g
Calculate the amount in moles (n)
=2.27
=4.54
Divide each of the moles by the smallest mole value in step 2
=1
=2
Obtain the simplest whole number mole ratio
Example 2
A sample of a compound of aluminium and oxygen has a mass of 2.36g. It contains 1.25g Al. Find the empirical formula0.0463
0.0694
1.0
1.5
Therefore the empirical formula is Al2O3
Calculating Molecular formula
Overview - calculate the empirical formula, then use the mass given to identify the number of molesExample
Hydrocarbon contains 7.2g of C and 1.5g of H.The molar mass of the compound was 58 g mol -1.
= 0.6
= 1.5
= 1
= 2.5
the molecular mass of C2H5 is 12 x 2 + 1 x 5 = 29g
But the molar mass of the compound is 58g ( data given from the question)
Number of C2H5 s in one of these 58g moles is 58 / 29 = 2
therefore the molecuar formula is 2 x the empirical formula
ie C4H10
Extension activity
http://phet.colorado.edu/en/simulation/isotopes-and-atomic-mass
http://phet.colorado.edu/en/simulation/molarity